First of all, every number in each row is odd, so we can eliminate 392 and 394.
The number of elements in each row forms an arithmetic sequence:
• 1st row : 1 element
• 2nd row : 3 elements - and 3 - 1 = 2
• 3rd row : 5 elements - and 5 - 3 = 2
• 4th row : 7 elements - and 7 - 5 = 2
and so on, so that the [tex]n[/tex]-th row has [tex]1 + 2(n-1) = 2n - 1[/tex] elements.
This means that in [tex]n-1[/tex] complete rows, there is a total of
[tex]\displaystyle \sum_{i=1}^{n-1} (2i-1) = n^2 - 2n + 1[/tex]
elements, so that the first element in the subsequent [tex]n[/tex]-th row is the [tex](n^2-2n+2)[/tex]-th number in the sequence {1, 3, 5, 7, …}, i.e the [tex](n^2-2n+2)[/tex]-th odd positive integer. To compute the sum, I use the following well-known formulas.
[tex]\displaystyle \sum_{i=1}^n 1 = n[/tex]
[tex]\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}2[/tex]
Now, the [tex]n[/tex]-th term of the sequence {1, 3, 5, 7, …} is simply [tex]2n-1[/tex], the [tex]n[/tex]-th odd positive integer. So the first element in the [tex]n[/tex]-th row is
[tex]2(n^2-2n+2) - 1 = 2n^2 - 4n + 3[/tex]
and hence the 1st element of the 15th row is
[tex]2\cdot15^2 - 4\cdot15 + 3 = \boxed{393}[/tex]
